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Thread: Trig Intervals

  1. #1
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    Trig Intervals

    I'm trying to figure out this problem. I think I have the answer but I'm not sure.

    Find all solutions on the interval [0,2$\displaystyle {\pi} $]. When appropriate put you answer in interval notation.

    $\displaystyle cos(x)-\frac{\sqrt{3}}{2}>0 $

    1) this is the same as $\displaystyle cos(x)+0>\frac{\sqrt{3}}{2} $ correct?
    2) I know that cosx is equal to $\displaystyle \frac{\sqrt{3}}{2} $ at $\displaystyle \frac{\pi}{6},\frac{11\pi}{6} $

    but, since it is asking for greater than, I graphed out cosx and thought that the answer is 0, $\displaystyle {2\pi} $. am I close?
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  2. #2
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    Hi discobob,

    What exactly does the graph of cos(x) look like from 0 to 360 degrees?

    It starts at cos(0)=1,
    drops to zero at 90 degrees,
    goes down to -1 at 180 degrees,
    comes back up to zero at 270 degrees,
    reaches 1 again at 360 degrees.

    Therefore there is an interval from zero to where cos(x) is $\displaystyle \frac{\sqrt{3}}{2}$

    where $\displaystyle cos(x)>\frac{\sqrt{3}}{2}$

    and similar at the other end of the period.
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  3. #3
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    yes, that is the graph i plotted. I think I dont understand how to figure out the interval where $\displaystyle cos(x)>\frac{\sqrt{3}}{2} $. I can see the area where it is greater than $\displaystyle \frac{\sqrt{3}}{2} $, but i dont know how to define it.

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  4. #4
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    Oh! I think I got it.

    Is the answer $\displaystyle [0,\frac{\pi}{6}) U (\frac{11\pi}{6},2\pi] $
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  5. #5
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    Note that you need to solve $\displaystyle \cos x-\cos \frac{\pi }{6}=2\sin \left( \frac{6x+\pi }{12} \right)\sin \left( \frac{\pi -6x}{12} \right)>0.$
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  6. #6
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    you've got it, discobob
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